Страница 244 номер 821, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\left(\frac{2x^{2} - 3x + 1}{x + 1}\right)' = \frac{(2x^{2} - 3x + 1)'(x + 1) - (2x^{2} - 3x + 1)(x + 1)'}{(x + 1)^{2}} =\)
\(= \dfrac{(4x - 3)(x + 1) - (2x^{2} - 3x + 1)}{(x + 1)^{2}} =\)
\(= \dfrac{4x^{2} + 4x - 3x - 3 - 2x^{2} + 3x - 1}{(x + 1)^{2}} = \dfrac{2x^{2} + 4x - 4}{(x + 1)^{2}};\)

2) \(\left(\frac{3x^{2} + 2x - 1}{2x + 1}\right)' = \frac{(3x^{2} + 2x - 1)'(2x + 1) - (3x^{2} + 2x - 1)(2x + 1)'}{(2x + 1)^{2}} =\)
\(= \dfrac{(6x + 2)(2x + 1) - 2(3x^{2} + 2x - 1)}{(2x + 1)^{2}} =\)
\(= \dfrac{12x^{2} + 6x + 4x + 2 - 6x^{2} - 4x + 2}{(2x + 1)^{2}} = \dfrac{6x^{2} + 6x + 4}{(2x + 1)^{2}};\)

3) \(\left(\frac{2 - x}{\sqrt{x}} + \frac{\sqrt{x}}{2 - x}\right)' = \frac{(2 - x)'\sqrt{x} - (2 - x)\left(\sqrt{x}\right)'}{x} + \frac{\left(\sqrt{x}\right)'(2 - x) - \sqrt{x}(2 - x)'}{(2 - x)^{2}} =\)
\(= \dfrac{-\sqrt{x} - \frac{2 - x}{2\sqrt{x}}}{x} + \dfrac{\frac{2 - x}{2\sqrt{x}} + \sqrt{x}}{(2 - x)^{2}} =\)
\(= \dfrac{\frac{-2x - 2 + x}{2\sqrt{x}}}{x} + \dfrac{\frac{2 - x + 2x}{2\sqrt{x}}}{(2 - x)^{2}} = \dfrac{-2 - x}{2x\sqrt{x}} + \dfrac{2 + x}{2(2 - x)^{2}\sqrt{x}} =\)
\(= \dfrac{1}{2\sqrt{x}}\left(\dfrac{-2 - x}{x} + \dfrac{2 + x}{(2 - x)^{2}}\right);\)